The following questions occurred to me.
This is not research mathematics, just idle curiosity.
Apologies if it is inappropriate.

Suppose you have a fixed volume V of maleable material,
perhaps clay.
The goal is to form it into a shape S (convex or nonconvex) that would roll down an inclined
plane as fast as possible.
The plane is tilted at θ with respect to the horizontal.
The race track that quantifies "as fast as possible"
is of length L. The shape S must be
entirely behind a starting plane orthogonal to the inclined plane,
and its race is finished when it is entirely ahead of a finishing plane,
L distant, again orthogonal to the inclined plane.
So if S is a disk of radius r, its center of
gravity will have to travel a total distance of L + 2r
to complete the race. S is released at rest, and rolls
under gravity. Assume that the materials of S and of the
inclined plane have a sufficient coefficient of friction μ
such that there is no slippage, just pure rolling.
The stipulation that S start behind and end
ahead of the orthogonal planes suggests that the optimal shape
is an arbitrarily thin (and therefore arbitrarily long) cylinder, to minimize r.
But this does not accord with (my) physical intuition.
What am I missing here?

A variation on the same problem replaces
the inclined plane with an inclined half-cylinder of radius R, like a rain gutter.
It may be necessary to assume a relation between R and the volume V of material
to make this problem reasonable (so that S fits in the gutter). But certainly this version's
answer could not be an arbitrarily long cylinder!

Joseph, can you, please, clarify whether you allow for non-convex S ? If so and if, moreover, S can "stick out of the gutter", a variant of the "pair of wheels" construction may solve problem 2 as well, but I don't want to rewrite the answer for nothing.
–
Victor ProtsakMay 26 '10 at 6:00

The problem with one less dimension seems interesting as well: the fixed volume imposes greater restriction, and one can expect the disc to be the optimal shape (though it seems not obvious to prove).
–
Benoît KloecknerMay 26 '10 at 10:33

Another remark: I guess you assume that the shape stays in contact with the plane, but this is not physically the case for all shapes and length. I guess any non-round cylindrical shape will jump after a long enough run.
–
Benoît KloecknerMay 26 '10 at 10:36

2 Answers
2

There is one thing that's way more important than the precise definition of how the race starts and ends: the moment of intertia of $S$, which determines how much energy is wasted on the rotation. Let us assume that $S$ is cylindrically symmetrical with mass $m$, radius $R$, moment of intertia $I$, and let $k=I/mR^2$. Then from the conservation of energy,

$$\frac{I\omega^2}{2}+\frac{mv^2}{2}=mgx\sin\theta,$$

and using that $v=\omega R,$ we get that

$$\frac{dx}{dt}=v=\sqrt{\frac{2g\sin\theta}{k+1}}x^{1/2}.$$

This already makes it clear that in order to minimize the rolling time, $k$ must be as small as possible, but this can also be quantified by solving the separable ODE and finding that

where $d=L+2R$ is the distance traveled. The conclusion, in the cylindrically symmetrical case of radius $R$, is that $S$ should be as close as possible to the infinitely thin heavy rod with a disk-shaped thin "flap" of radius $R$ attached, so that $k$ is just slightly over $1$. In a practical sense, imagine a wheel pair with very light rims and a very heavy axle. [I like to think of this situation as "anti-flywheel problem": we want to minimize rotation energy accumulated by $S$, so contrary to the usual flywheel, the wheel is light and the axle is heavy.] If your "malleable material" has uniform density and $R$ may vary, one sequence of mathematical solutions approaching the time bound consists of solid cylinders with radii $r_i$ shrinking to 0 and growing widths $\ell_i$, with two $w$-thin disks of radius $R_i$ attached, where

$$\pi(\ell_i r_{i}^2+2w R_i^2)=V, \quad r_i=o(R_i), \quad R_i\to 0.$$

[I've only put in an explicit sequence because of a clarification request in the comments; I actually think that this level of detail hinders understanding]. It's also clear that the cylindrical symmetry restriction is irrelevant.

The second problem is more difficult: I don't have much to say on it other than that, by the same token, a solid ball is better than a disk/cylinder even if it could be fitted in the "gutter".

P.S.: "Physical" intuition may be deceptive: many people assume that a ball will roll according to the same law of motion as a cylinder or an ideal point mass. This actually happened to some of my mathematically oriented classmates in a physics lab assignment, who tried to make up data in order to avoid doing tedious experiments.

Is the cylindrical symmetry really irrelevant? For a fixed width the heavy rod is the best solution. But for a fixed maximum radius, your wheel pair with heavy axle is better. Since the inf lies in a non-compact direction, there may be multiple, not necessary the same methods of finding a minimizing sequence. In fact, it is probably also possible to construct a one-parameter family of shapes such that a stationary point (relative to the parameter) is achieved.
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Willie WongMay 25 '10 at 23:35

Since there was no width restriction in problem 1, the answer is yes, cylindrical symmetry is irrelevant. The point is that you want to waste as little energy as possible on rotation. However, you've made a good point. Re-reading my answer, I see that I was a bit sloppy with the assumptions: when I said "heavy rod", I wasn't yet thinking of uniform density. Let me mull over it a bit more.
–
Victor ProtsakMay 26 '10 at 0:03

What does $I$ stand for in your formula, $k=I/mR^2$?
–
Gerry MyersonMay 26 '10 at 0:17

I've put in some clarifications, but it's supposed to be a fun problem, you can work out the details yourselves!
–
Victor ProtsakMay 26 '10 at 1:15

2

Joseph, suppose you just take a cylinder of uniform density. If you make the radius smaller, the moment of inertia gets smaller, but the cylinder also has a smaller perimeter, so it has to rotate more to get to the bottom. These two effects cancel out -- it doesn't matter how large you make the cylinder, 1/3 of the energy goes into spinning the cylinder and 2/3 into rolling it. If the body has to be convex and circularly symmetric, the best shape is presumably a thin cylinder with two circular pyramids glued onto either side. Without the convexity restriction, Victor's analysis works
–
Peter ShorMay 26 '10 at 21:03

I am going to post a bunch of assumptions which will impose further constraints on the 2nd problem. I think these assumptions maybe natural in view of the thought experiment.

Without loss of generality, we can assume that the gutter has width 1, so the width of the gutter is parametrized by $[0,1]$.

The cross-section of the gutter itself is parametrized by some convex function $g: [0,1] \to \mathbb{R}$ with $g(0) = g(1)$.

The object is not allowed to stick out of the gutter side-ways (else we can just take a heavy rod that rests on the top of the gutter...).

The object is assumed to have uniform density and total mass 1.

The object is obtained by rotation (since it is to rotate) around an axis.

First, by energy conservation. similar to described in Victor Protsak's answer, to minimize the total time of travel it suffices to minimize
$$ \mathcal{E} = (L + 2 R_0)( 1 + I / R_1^2 ) $$
where $R_0$ is the maximum radius of the rotating body and $R_1$ is radius at which it is rolling, in other words the maximum distance from the axis to a point on the surface of the object that is in contact with the gutter. (Note that it is impossible for a ball of radius 1 to roll inside a gutter of radius 1 without slipping!!! We'll just conveniently forget frictional dissipation for our computations.)

Now, by the definition of the angular momentum and simple re-arrangement argument, it is clear that the cross-section of the object can be obtained as a function over its axis. More precisely, in a cylindrical coordinate system where $z$ is along the axis of rotation, the object
$$ S = \{ r \leq f(z) \} $$
(If there are any "holes" near the axis, by moving the mass inward closer to the axis it reduces both $I$ and $R$.)

It remains to write $I$ and $R$ as a functional of $f$. Working in the cylindrical coordinate still, the volume of $f$ is
$$ V(f) = \pi \int_0^1 f(z)^2 dz $$
By assumption this is a constant number.

Lastly, $R_0 = \sup f$ by definition. But $R_1$ is a bit tricky to compute. A bit of thought tells you that $R_1 = \sup_{P} f(z)$, where $P$ is the set of points satisfying $f(z) + g(z) = \sup (f+g)$.

Observe that $R_1 \leq R_0$.

To summarize, the task then becomes: given a function $g(z)$ and constants $\alpha,\beta,V$, minimize the functional
$$\mathcal{E} = (\alpha + \|f\|_\infty)(\beta + \frac{\|f\|^4_4}{R_1(f;g)^2} )$$
under the constraint that $\|f\|_2^2 = V$, where $\|\cdot\|_p$ denotes the Lebesgue $p$-norm over $[0,1]$.

If we impose the constraint that $R_1 = R_0$, then the problem reduces to the first one which Victor examined, where the solution is a Rolling cylinder. I rather suspect that the cylinder is also the solution in the general case. (I am pretty sure that to minimize $L^4$ while fixing $L^2$ on a compact interval one needs to make the function as flat as possible.)